Find $$\lim_{n\to\infty}\frac{a}{n} \left(1 + \sqrt{\frac{n}{n+a}} + \sqrt{\frac{n}{n+2a}} + \cdots + \sqrt{\frac{n}{n+na}}\right)$$
What I have done by now to answer:
First I did a little bit of programming to figure out what the answer is maybe it give me a sense of what I should do I arrived at $2(\sqrt{a+1} - 1).$
And the other thing that come to my mind is that difference between each term of summation is about $0$ and as there are infinite terms I can write the summation as an integral of square root function between $1/(a+1)$ and $1$.
$$\begin{array}{lcl} \displaystyle \frac{a}{n} \left(1 + \sqrt{\frac{n}{n+a}} + \sqrt{\frac{n}{n+2a}} + \cdots + \sqrt{\frac{n}{n+na}}\right) & = & \displaystyle \dfrac{a}{n} \sum_{k = 0}^n \dfrac{1}{\sqrt{1 + a \dfrac{k}{n}}} \\[5mm] & \to & \displaystyle a \int_0^1 \dfrac{\mathrm{d}x}{\sqrt{1 + a x}} \\[5mm] & = & \displaystyle 2 \left[\sqrt{1 + a x}\right]_0^1 \\[5mm] & = & 2 \sqrt{1 + a} - 2 \end{array}$$